Question

10. A boiler system is being designed for a building expected to require heating for approximately 3,500 hours per year. The LDC is indicated below (left). The total annual heating load is about 280,000 kWh. The load will be met by three identical boilers, with each boiler having the following characteristics: • Heating Fuel: natural gas (heat content may be estimated to be = 10.33 kWh/mº). • Burner Type: on/off • Rated Output = 100 kW Rated Efficiency: na= 82% The boiler operating sequence will be as follows: Boiler-1 fires first and operates to meet whatever load it can, up to its capacity. Boiler 2 fires next and operates to meet whatever load it can (above that met by Boiler-1), up to its capacity. Boiler-3 is then added sequence in a similar way to that of Boiler-2. The average heat output of each boiler will be adjusted by cycling its burner on and off. The hourly part-load performance of each boiler can be estimated using the adjustment curve below (right). Load Duration Curve Boiler Performance Adjustment Curve 300 250 100% 200 90% Heating Load [kW] NPL 80% 150 FL 100 70% 50 60% 0 50% 0% 20% 40% 60% 80% 100% 0 3000 1000 2000 Hour Number PLR Based on the information provided above, make reasonable estimates of the following: i) annual heating load met by Boiler-1 ii) average efficiency of the three boilers when operating on the coldest day of the year.

Answer 1

son Por sching the question are have to look at the graph very casifully Given Data, Bunnel tye: LEX ?? un bu Heat content and feel = CV = 10.33 kwh/m3 con ( OFF Pated output = p = 100 Ich / Boiles , Rated efficiency (Full load 282% offienry) = 2 As we know that boiler will run at full load effiving load requirement is above 100 kw. for to 1500 hrs. Pout, => for 750 hrs to 2500 hrs. I Porut <$100 kW) → As that we can see that from Right side curse at PLR 7, 50% Me >95% EL Npc =( 0.95*0.82) { for safe working mpu 00479 100 kW

> for 2500 has to 3500 hrs. (Pout <50kw) "UpL = 80.7. & estimated from light graphi RE - Фрі 0.880.82 0-656 sos Annual boiler -(approx) heal met by (Qi Jonume) 100 ( 790-0) + 50 (2500-450) +4x (116-50) * (2500-750) + 4x50 x ( 350-2500 (Qi Jannical z 231250 kWh Ans: ©). coldest day At all the boite will be fully operated at total theating load of. 250 Kloßr, > Boiler-1 s Boiler -2 ore at full load but Boiles -3 will be at part Mp) DEL /Boile-3 load of 50% 0-95 play 0.95 x 0.82 = 0.779

SO average (972), + (2.p.), + (npu) 3 efforening of - three boiles 0.82 7 0.82 +0.779 3 0.80633 or 80.633% Mu fy

Answer may slightly vary from person to person as these are the estimates . Please forgive me for any calculation mistake and do comment for any short of queries.